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kykrilka [37]
3 years ago
11

GIVING BRAINLIEST :D

Chemistry
1 answer:
horrorfan [7]3 years ago
4 0

Answer:

0.1 M molar NaaOH

Explanation:

M1V1= M2V2

M1 = 0.150M NaOH

V1=100 ml

V2=150 ml

M2=?

M2= M1V1/V2

M2= 0.150*100/150=0.1 M

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For the reaction H2PO4- HAsO4 2-HPO4 2-+H2AsO4- <br> what species are a conjugate acid-base pair?
aleksklad [387]

Ionic equation:

\text{H}_2\text{PO}_4^{-} + \text{HAsO}_4^{2-} \to \text{HPO}_4^{2-} + \text{H}_2\text{AsO}_4^{-}

The acid and base in a conjugate pair differ by only one proton \text{H}^{+}. The acid loses one proton to produce a conjugate base, whereas the base gains a proton to produce its conjugate acid.

\text{H}_2\text{PO}_4^{-} loses one proton to produce \text{HPO}_4^{2-} in this reaction.

\text{H}_2\text{PO}_4^{-} \to \text{H}^{+} + \text{HPO}_4^{2-}

Meanwhile, \text{HAsO}_4^{2-} gains one proton to form \text{H}_2\text{AsO}_4^{-}.

\text{HAsO}_4^{2-} + \text{H}^{+} \to \text{H}_2\text{AsO}_4^{-}

Therefore

  • \text{H}_2\text{PO}_4^{-} is the conjugate acid  \text{HPO}_4^{2-}, its conjugate base.
  • \text{HAsO}_4^{2-} is the conjugate base of \text{H}_2\text{AsO}_4^{-}, its conjugate acid.
8 0
3 years ago
Read 2 more answers
What is the mole fraction of each component if 3.9 g of benzene (C6H6) is dissolved in 4.6 g of toluene (C7H8)
Savatey [412]

Answer:

Step 1 of 6

(a)

The mass of benzene is  , so calculate the moles of benzene as follows:



The mass of toluene is, so calculate the moles of toluene as follows:



Now, calculate the mole fraction as follows:





Therefore, the mole fraction of benzene and toluene is  and  respectively.

Step 2 of 6

(b)

The formula to calculate the partial pressure is as follows:



Here,  is the partial pressure of benzene,  is the vapour pressure of pure benzene and  is the mole fraction of benzene.

Vapour pressure of pure benzene at  is.

Substitute the values in the equation as follows:



Therefore, the partial pressure is  .

Step 3 of 6

(c)

Vapor pressure of the solution at 1 atm is  .

When the total pressure of the vapour pressure of the mixture is  at a temperature, then, the solution boils. It corresponds to the boiling point of the solution.

Calculate the total pressure of the solution at  as follows:



Since, the total pressure is less than the atmospheric pressure, the solution will not boil at  .

Calculate the total pressure of the solution at  as follows:



Since, the total pressure is greater than the atmospheric pressure, the solution will boil at  .

Therefore, the boiling point of the solution is  .

Step 4 of 6

(d)

Mole fraction of benzene at  is calculated as follows:



Mole fraction of toluene at  is calculated as follows:



Therefore, the mole fractions of benzene and toluene are  and  respectively.

Step 5 of 6

(e)

Vapor pressure of benzene at  is  .

Partial pressure of benzene is calculated as follows:



Vapor pressure of toluene at  is  .

Partial pressure of toluene is calculated as follows:



Step 6 of 6

Weight composition of the vapour that is in equilibrium with the solution is calculated as follows:



Weight composition of the vapour that is in equilibrium with the solution is calculated as follows:



Explanation:

mark me as brainliest

4 0
3 years ago
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Which statement best describes an array of 10 elements where the elements are of a class type?
ladessa [460]
The array will be created with the elements equal to null.
3 0
3 years ago
Need help <br>1/6 + 1/8 =
pogonyaev

1/6 + 1/8 = 7/24

24

≅ 0.2916667

6 0
3 years ago
Calculate the pH of a buffer solution prepared by mixing 60.0 mL of 1.00 M lactic acid and 25.0 mL of 1.00 M sodium lactate.
marshall27 [118]
This problem could be solved easily using the Henderson-Hasselbach equation used for preparing buffer solutions. The equation is written below:

pH = pKa + log[(salt/acid]

Where salt represents the molarity of salt (sodium lactate), while acid is the molarity of acid (lactic acid). 

Moles of salt = 1 mol/L * 25 mL * 1 L/1000 mL = 0.025 moles salt
Moles of acid = 1 mol/L* 60 mL * 1 L/1000 mL = 0.06 moles acid
Total Volume = (25 mL + 60 mL)*(1 L/1000 mL) = 0.085 L

Molarity of salt = 0.025 mol/0.085 L = 0.29412 M
Molarity of acid = 0.06 mol/0.085 L = 0.70588 M

Thus,
pH = 3.86 + log(0.29412/0.70588)
pH = 3.48
4 0
3 years ago
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